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Consider $X_1,X_2,...,X_n$ i.i.d $U(-\theta,0)$.

I want to find the maximum likelihood estimator of $\theta$.

I know that $f(x,\theta)=\frac{1}{\theta}$ for $-\theta < x < 0$ and that $L_n(\theta, x)= \frac{1}{\theta^n}$.

If we were looking at $U(0,\theta)$, then the MLE of $\theta$ would be $x_{(n)}$ because $L_n(\theta, x)= \frac{1}{\theta^n}$ is decreasing from $0 < x < \theta$ and would thus be maximized at the max $x_i$, which is $x_{(n)}$.

For my case, since $L_n(\theta, x)= \frac{1}{\theta^n}$ is an increasing function for $-\theta < x < 0$, then $L_n(\theta, x)= \frac{1}{\theta^n}$ will be maximized at the max $x_i$, and thus the MLE of $\theta$ will be $x_{(n)}$ as well.

I think this is correct, but it seems very silly to me that for both cases you can just say that it will be maximized at the max $x_i$. Could someone better explain this to me?

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    $\begingroup$ How is the likelihood increasing or decreasing? They look constant in $x$. $\endgroup$ – jdods Dec 15 '17 at 3:52
  • $\begingroup$ You're right! Strange, in my notes I have a graph of $1/\theta^n$ with $\theta$ on the x-axis, maybe it was referring to something else @jdods $\endgroup$ – Silvia Rossi Dec 15 '17 at 4:00
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    $\begingroup$ If max gives you info on right end of interval for $Unif(0,\theta),$ then what would give you info on left end for $Unif(-\theta,0)?$ (You are on the right track that this is not a maximization problem with differentiation.) $\endgroup$ – BruceET Dec 15 '17 at 4:11
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Note that the likelihood function is a function of $\theta$. In particular, $$L_n(\theta;\vec X) = \left \{ \begin{matrix}\frac{1}{\theta^n} & \text{if $\theta \ge -X_i$ for $i=1,2,\cdots, n$,} \\ 0 & \text{otherwise.}\end{matrix}\right.$$ Here $\theta \ge -X_i$ comes from $-\theta \le X_i$. Now, $L_n(\theta,\vec X)$ is a decreasing function of $\theta$. Consequently, $L_n(\theta;\vec x)$ attains its maximum when $\theta = \max\{-X_i\}=-\min\{X_i\}=-X_{(1)}$.

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Note that all $X_i$ are negative, but you may want $\theta$ to be positive. So you cannot choose $X_{(n)}$ as the MLE because that would give a negative MLE.

If $-\theta > X_{(1)} $, then the likelihood will be zero because $X_{(1)}$ is not in the support of the distribution. If $-\theta \leq X_{(1)} $, then the likelihood will be $\frac{1}{\theta^n}$. So the maximum likelihood is attained at $\theta = - X_{(1)}$.

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Your reasoning for the $U(0,\theta)$ case is wrong, so is interfering with the $U(-\theta,0)$ case.

In the $U0,\theta)$ case the likelihood (which is a function of $\theta$ !) is $\frac{1}{\theta^n}$ if $\theta\ge x_{(n)}$ and zero otherwise since the support requires all $x_i <\theta.$ Since $1/\theta^n$ is decreasing in $\theta,$ you must pick the smallest value of $\theta$ that is $\ge x_{(n)},$ i.e. $x_{(n)}.$

Likewise for $U(-\theta,0),$ the likelihood is $1/\theta^n$ for $\theta \ge -x_{(1)}$ and zero otherwise since the support requires all $x_i \ge -\theta.$ So the MLE is $-x_{(1)}.$ This should make sense since this is just a mirror image of the previous problem.

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